The latent heat of vaporization of H2O at body temperature (37°C) is 2.42 x 106 J/kg. To cool the body of a 77.6-kg jogger [average specific heat capacity = 3500 J/(kg C°)] by 2.78 C°, how many kilograms of water in the form of sweat have to be evaporated?
(77.6)(3500)(2.78)=M(2.42E^6)
755048=M(2.42E^6)
M=0.312
Well, I guess the jogger must be sweating like a marathon runner in a sauna! Let's calculate how many kilograms of water need to be evaporated to cool them down.
First, we need to figure out the amount of heat needed to cool down the jogger. We can use the formula:
Q = mcΔT
where Q is the amount of heat, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
Q = (77.6 kg) * (3500 J/(kg °C)) * (2.78 °C)
Q ≈ 755,872 J
Now, let's convert this amount of heat into the mass of water that needs to evaporate. For that, we'll use the formula:
Q = mL
where Q is the amount of heat, m is the mass of water, and L is the latent heat of vaporization.
m = Q / L
m = 755,872 J / (2.42 × 10^6 J/kg)
m ≈ 0.312 kg
So, our sweaty jogger needs to lose approximately 0.312 kilograms of water in the form of sweat to cool down. But don't worry, they'll rehydrate soon enough!
To calculate the amount of water that needs to be evaporated to cool the body of the jogger, we need to find the heat required to decrease the body temperature by 2.78°C.
The formula for heat transfer is:
Q = mcΔT
Where:
Q = heat transfer (in joules)
m = mass (in kilograms)
c = specific heat capacity (in joules per kilogram per degree Celsius)
ΔT = change in temperature (in degrees Celsius)
Let's plug in the given values:
m = 77.6 kg (mass of the jogger)
c = 3500 J/(kg·°C) (specific heat capacity of the jogger)
ΔT = -2.78°C (change in temperature, negative because we are cooling the body)
Q = (77.6 kg) × (3500 J/(kg·°C)) × (-2.78°C)
Q = -762,944 J
The negative sign indicates that heat is being transferred out of the body.
Now, we know that the heat required to evaporate water is equal to the product of the mass of water (in kg) and the latent heat of vaporization.
The formula for heat transfer due to evaporation is:
Q = mL
Where:
Q = heat transfer (in joules)
m = mass (in kilograms)
L = specific latent heat of vaporization (in joules per kilogram)
Let's rearrange the equation to solve for the mass of water:
m = Q / L
Plugging in the values:
Q = -762,944 J (heat required to cool the body)
L = 2.42 × 10^6 J/kg (latent heat of vaporization of water at body temperature)
m = (-762,944 J) / (2.42 × 10^6 J/kg)
m ≈ -0.315 kg
The negative sign indicates that the mass of water is lost due to evaporation. However, mass cannot be negative, so we can interpret it as the absolute value.
Therefore, approximately 0.315 kg (or 315 grams) of water needs to be evaporated in the form of sweat to cool the body of the jogger by 2.78°C.
To solve this problem, we need to understand the concept of heat transfer. The heat required to change the temperature of a substance can be calculated using the specific heat capacity, and the heat required to change the phase of a substance (like from liquid to gas) can be calculated using the latent heat of vaporization.
Here are the steps to find the amount of water that needs to be evaporated:
1. Calculate the heat required to cool the jogger's body:
Heat = mass * specific heat capacity * change in temperature
mass = 77.6 kg (given)
specific heat capacity = 3500 J/(kg °C) (given)
change in temperature = 2.78 °C (given)
Heat = 77.6 kg * 3500 J/(kg °C) * 2.78 °C
Heat = 76,951.6 J
2. Calculate the mass of water to be evaporated:
Heat = mass * latent heat of vaporization
latent heat of vaporization = 2.42 x 10^6 J/kg (given)
76,951.6 J = mass * 2.42 x 10^6 J/kg
mass = 76,951.6 J / (2.42 x 10^6 J/kg)
mass = 0.0318 kg
Therefore, approximately 0.0318 kg (or 31.8 grams) of water in the form of sweat needs to be evaporated to cool the body of the jogger by 2.78 °C.